Is that a homework? If so, it should be tagged as such.
–
mbqDec 20 '11 at 13:48

I think the same issue would manifest itself with two dense clouds of particles moving together, each at speed $v$ with respect to some observer, and colliding in the observer's reference frame, but it might be easier to think about that way.
–
David Z♦Dec 20 '11 at 19:17

1 Answer
1

The first answer is correct. The second one is wrong because it ignores the energy of the heavy body.

In your example, imagine some putty hitting a wall. The putty is moving to the right and the wall is stationary. When there's a collision, the wall picks up some very tiny velocity $w$. That velocity is basically inversely-proportional to the wall's mass $M$. The kinetic energy it picks up is proportional to $Mw^2$, and thus also inversely proportional to $M$. Therefore, in this frame the wall's energy is negligible.

However, if there is a man moving to the left, he sees the putty moving faster, as you say, but he also sees the wall moving to the right. When the putty hits it, the wall will move a tiny bit faster to the right and pick up a little bit more kinetic energy. This time, though, the wall's energy gain cannot be made negligible by making the wall more massive. You can work out easily that in the limit $M\to\infty$ the kinetic energy the wall gains is $mvV$, compensating for that lost by the putty in this frame.

Here is a general solution. The kinetic energy of a particle is

$$T = \frac{\mathbf{p}^2}{2m}$$

If we boost into a frame moving at speed $\mathbf{v}$ relative to the original, the new momentum is

which is exactly the same as the expression for a single particle of momentum $\mathbf{p}_1 + \mathbf{p}_2$ and mass $m_1 + m_2$.

Say you are watching two particles. The energy is $E_1$. Before they collide, you boost to a new frame. In that new frame, the new kinetic energy is 100J greater, so the energy is $E_1 + 100J$. Then the particles collide. The energy becomes $E_2$.

You could also have waited for the particles to collide first, then boosted to a new frame. In that case, the energy after the collision and before the boost is $E_f$. What we showed above is that the boost still results in a gain of 100J, so the energy after the boost is $E_f + 100J$.

It is evident that $E_2 = E_f + 100J$ because they are the same physical situation. Therefore

$$E_2 - (E_1 + 100J) = E_f + 100J - (E_1 + 100J) = E_f - E_1$$

The left hand side is the energy lost during the collision in the boosted frame. The right hand side is the energy lost during the collision in the original frame. Thus, when there is an inelastic collision going on, boosting to a new frame only changes the total amount of kinetic energy around. It does not change the amount of kinetic energy transformed to thermal energy.